Question: Simplify and expand the following expression: $ \dfrac{5z}{2z - 9}-\dfrac{4z}{z + 3} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2z - 9)(z + 3)$ Multiply the first term by $\dfrac{z + 3}{z + 3}$ $ \begin{align*} \dfrac{5z}{2z - 9} \times \dfrac{z + 3}{z + 3} & = \dfrac{(5z)(z + 3)}{(2z - 9)(z + 3)} \\ & = \dfrac{5z^2 + 15z}{(2z - 9)(z + 3)}\end{align*} $ Multiply the second term by $\dfrac{2z - 9}{2z - 9}$ $ \begin{align*} \dfrac{4z}{z + 3} \times \dfrac{2z - 9}{2z - 9} & = \dfrac{(4z)(2z - 9)}{(z + 3)(2z - 9)} \\ & = \dfrac{8z^2 - 36z}{(z + 3)(2z - 9)}\end{align*} $ Now we have: $ = \dfrac{5z^2 + 15z}{(2z - 9)(z + 3)} - \dfrac{8z^2 - 36z}{(z + 3)(2z - 9)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5z^2 + 15z - (8z^2 - 36z)}{(2z - 9)(z + 3)} $ $ = \dfrac{5z^2 + 15z - 8z^2 + 36z}{(2z - 9)(z + 3)} $ $ = \dfrac{-3z^2 + 51z}{(2z - 9)(z + 3)}$ Expand the denominator: $ = \dfrac{-3z^2 + 51z}{2z^2 - 3z - 27}$